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Boundary conditions for eigenfunctions in a potential step

  1. Sep 3, 2017 #1
    1. The problem statement, all variables and given/known data
    A particle with mass m and spin 1/2, it is subject in a spherical potencial step with height ##V_0##.
    How is the general form for the eigenfunctions?
    What is the boundary conditions for this eigenfunctions?
    Find the degeneracy level for the energy, when it is ##E<V_0##

    2. Relevant equations
    Radial equation
    \begin{equation}
    -\frac{\hbar^2}{2m}\left(\frac{\partial^2u}{\partial r^2}-\frac{l(l+1)}{r^2}u\right)+V(r)u(r)=Eu(r)
    \end{equation}
    with ##u(r)=rR(r)##


    3. The attempt at a solution
    Well, first the constant of motion it ##\hat{L_z}##, ##\hat{L^2}## and ##\hat{S_z}## because conmute with ##\hat{H}##, also ##\hat{H}##, and it former a complete set of observables that conmute, therefore the form of eigenfunctions are:
    \begin{equation}
    \psi(r,\theta,\varphi,\sigma)=R(r)Y(\theta,\varphi)\chi(\sigma)
    \end{equation}
    when ##\chi(\sigma)## is a spin function, and only take the values ##m_s={+1/2,-1/2}## because eigenvalues of operator ##\hat{S^2}## is ##3/4\hbar^2## with ##s=1/2##.

    Here my doubt:
    Boundary conditions to ##Y(\theta,\varphi)## this is referred to:
    ##l=0, 1, \dots, n-1## and ##-l\leq m\leq +l##?

    For radial function, it must be:
    because ##\hat{H}## is hermitian, then for ##r=0\Longrightarrow u(0)\longrightarrow0##
    and for ##r\longrightarrow\infty \Longrightarrow u\longrightarrow0##

    For ##\chi## the boundary conditions it is ##-s\leq m_s\leq+s##?

    In the case for the degeneracy, when ##E<0## corresponds to ##l=0## because it is a bound state, then without spin the level of degeneracy is ##n^2## but with spin is ##2n^2##.

    Up to here my poor attemp to solution, i am really puzzled for explain the boundary condition to Y and ##\chi##, and i am not sure that the
    ##r\longrightarrow\infty \Longrightarrow u\longrightarrow0## for this potential.

    Thanks a lot!
     
  2. jcsd
  3. Sep 8, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
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